1) Find the equation of the circle of radius $2$ with center at $(3, 0)$.
My answer: $\sqrt{(x-3)^2 + y^2} = 2$
2) Find the equation of the circle of radius $\sqrt3$ with center at (-1, -2).
My answer: $\sqrt{(x+1)^2 + (y+2)^2} = \sqrt3$
3) There are two circles of radius 2 that have centers on the line $x = 1$ and pass through the origin. Find their equations.
My answers: $\sqrt {(x-1)^2 + (y+\sqrt3)^2 }= 2$, $\sqrt {(x-1)^2 + (y-\sqrt3)^2 }= 2$.
4) Find the equation of the circle that passes through three points , $(0, 0)$, $(0, 1)$, $(2, 0)$.
My answers: The three points $(0, 1)$, $(0, 0)$, $(2, 0)$ make a right-angled triangle at $(0, 0)$. According to the Thales' theorem, the hypotenuse of the triangle which is a line-segment from the point $(2, 0)$ to $(0, 1)$ is the diameter of the circle. The center "P" lies on the point $(0+2/2, 1+0/2)$ = $P(2, 1/2)$. Therefore, the circle will be the locus of the equation: $\sqrt {(x-1)^2 + (y-1/2)^2} = r$
5) Find the equation of the circle one of whose diameters is the line segment from $(-1, 0)$ to $(5, 8)$.
My answers: $\sqrt {(x-2)^2 + (y-4)^2} = 5$
If any of the answers is wrong, so tell me the correct one please.
2) $\sqrt{(x+1)^2+(y+2)}=\sqrt 3$ (You should not replace $\sqrt 3$ by $1.73$)
3) $(S_1): (x-1)^2+(y-\sqrt 3)^2=4$ and $(S_2): (x-1)^2+(y+\sqrt 3)^2=4$
(Hint. The center is the intersecrion of the circle with origin center, radius=$2$ and the line $x=1$. There are two intersection points $I_1=(1,\sqrt 3)$ and $I_2=(1, -\sqrt 3)$)
4) $(S): (x-1)^2+(y-\frac{1}{2})^2=\frac 5 4$
(Hint. Three points make a rectangle triangle at (0, 0). Hence, Radius $I=(\frac{0+2}{2},\frac{1+0}{2})=(1, \frac 1 2)$, Radius $R=\frac{\sqrt{(0-2)^2+(1-0)^2}}{2}=\frac{\sqrt {5}}{2}$))
5) $(S): (x-2)^2+(y-4)^2= 25$
(Hint. Center $I=(\frac{-1+5}{2},\frac{0+8}{2})=(2, 4)$, Radius $R=\frac{d}{2}=\frac{\sqrt{(5-(-1))^2+(8-0)^2}}{2}=5$)